2.microprocessor microcontroller lab 1

7/26/2019 2.Microprocessor Microcontroller Lab 1

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CHENDU

COLLEGEOFENGINEERINGANDTECHNOLOGY

MADURANTAKAM.

KANCHEEPURAM-603311.

CS6412 MICROPROCESSOR ANDMICROCONTROLLER

LABORATORY MANUAL

Prepared byN.S.SHINIJA, ASSISTANT PROFESSOR

Faculty of Electronics & Communication Engineering

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CS6412MICROPROCESSOR AND MICROCONTROLLER LAB DEPT OF CSE

2

EXPT.NO NAME OF THE EXPERIMENT PAGENO

1 Basic arithmetic and Logical operations 3

2 Move a Data Block Without Overlap 10

3 Code conversion, decimal arithmetic and Matrixoperations.

11

4 Floating point operations, string manipulations, sortingand searching

17

5 Password Checking, Print Ram Size And System Date 31

6 Counters and Time Delay 36

7 Traffic light control 38

8 Stepper motor control 40

9 Digital clock 42

10 Key board and Display 45

11 Printer status 48

12 Serial interface and Parallel interface 49

13 A/D and D/A interface and Waveform Generation 53

14 Basic arithmetic and Logical operations 59

15 Square and Cube program, Find 2s complement of a

number70

16 Unpacked BCD to ASCII 72

INDEX




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EXPT NO: 01 DATE:

AIM:To write an Assembly Language Program (ALP) for performing the Arithmetic

operation of two byte numbers.

APPARATUS REQUIRED:

SL.NO

ITEM SPECIFICATION QUANTITY

1. Microprocessor kit 8086 kit 1

2. Power Supply +5 V dc 1

PROBLEM STATEMENT:Write an ALP in 8086 to add and subtract two byte numbers stored in the

memory location 1000H to 1003H and store the result in the memory location1004H to 1005H.Also provide an instruction in the above program to consider thecarry also and store the carry in the memory location 1006H.

ALGORITHM:

(i) 16-bit addition

Initialize the MSBs of sum to 0

Get the first number.

Add the second number to the firstnumber.

If there is any carry, increment

MSBs of sum by 1.

Store LSBs of sum.

Store MSBs of sum.

(ii) 16-bit subtraction

Initialize the MSBs of difference to0

Get the first number

Subtract the second number fromthe first number.

If there is any borrow, incrementMSBs of difference by 1.

Store LSBs of differenceStore MSBs of difference.

(iii) Multiplication of 16-bit

numbers:

Get the multiplier. Get the multiplicand

Initialize the product to 0.

Product = product +multiplicand

Decrement the multiplier by 1

If multiplicand is not equal to0,repeat from step (d) otherwisestore the product.

(iv)

Division of 16-bit numbers.

Get the dividend

Get the divisor

Initialize the quotient to 0. Dividend = dividenddivisor

If the divisor is greater, store thequotient. Go to step g.

If dividend is greater, quotient= quotient + 1. Repeat fromstep (d)Store the dividend

value as remainder.

BASIC ARITHMETIC AND LOGICAL OPERATIONS USING 8086

PROGRAMMING




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FLOWCHART

ADDITION SUBTRACTION

START

SET UP COUNTER (CY)

GET SECOND OPERAND

TO A

A = A + B

STORE THE SUM

START

GET FIRST

OPERAND TO A

SUBTRACT

SECOND OPERANDFROM MEMORY

STORE THE

DIFFERENCE

STOP

IS THEREANY CARRY

GET FIRST OPERAND

COUNTER =

COUNTER + 1

STORE THE CARRY

STOP

SET UP COUNTER (CARRY)

IS THEREANY CY

COUNTER =

COUNTER + 1

STORE THE CARRY

NO

YES

YES

NO




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ADDITION

ADDRESS Opcodes PROGRAM COMMENTS

MOV CX, 0000H Initialize counter CX

MOV AX,[1200] Get the first data in AX reg

MOV BX, [1202] Get the second data in BX reg

ADD AX,BX Add the contents of both theregs AX & BX

JNC L1 Check for carry

INC CX If carry exists, increment theCX

L1 : MOV [1206],CX Store the carry

MOV [1204], AX Store the sum

HLT Stop the program

SUBTRACTION

ADDRESS OPCODES PROGRAM COMMENTS

MOV CX, 0000H Initialize counter CX

MOV AX,[1200] Get the first data in AX reg

MOV BX, [1202] Get the second data in BX reg

SUB AX,BX Subtract the contents of BX fromAX

JNC L1 Check for borrow

INC CX If borrow exists, increment theCX

L1 : MOV [1206],CX Store the borrow

MOV [1204], AX Store the difference

HLT Stop the program




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RESULT:.

ADDITION

MEMORY

DATA

SUBTRACTION

MEMORY

DATA

MANUAL CALCULATION

.




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FLOWCHART

MULTIPLICATION DIVISION

Start

Get Multiplier & Multiplicand

MULTIPLICAND

REGISTER=00

REGISTER =REGISTER +

MULTIPLICAND

Multiplier=MU

LTIPLIER1

MULTIPLIER

IS

=0?

NO

YES

Start

Load Divisor &

Dividend

QUOTIENT = 0

DIVIDEND =

DIVIDEND-DIVISOR

QUOTIENT = QUOTIENT+1

DIVIDEND BL go to L1

DEC SI Else, Decrement the memory location

MOV [SI],AL Store the largest data

MOV AL,BL Get the next data AL

JMP L2 Jump to L2

L1 : DEC SI Decrement the memory location

MOV [SI],BL Store the smallest data in memorylocation

L2 : INC SI Go to next memory location

DEC DL Decrement the count

JNZ L3 Jump to L3, if the count is not reachedzero

MOV [SI],AL Store data in memory location

DEC CL Decrement the count

JNZ L4 Jump to L4, if the count is not reachedzero

HLT Stop

RESULT:.

ASCENDING

MEMORY

DATA

DESCENDING

MEMORY

DATA

Thus given array of numbers are sorted in ascending & descending order.




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LARGEST& SMALLEST

AIM:

To write an Assembly Language Program (ALP) to find the largest and

smallest number in a given array.

APPARATUS REQUIRED:

SL.NO


1. Microprocessor kit 8086 1


PROBLEM STATEMENT:

An array of length 10 is given from the location. Find the largest andsmallest number and store the result.

ALGORITHM:

(i) Finding largest number:

a. Load the array count in a register C1.b. Get the first two numbers.c.

Compare the numbers and exchange if the number is small.d. Get the third number from the array and repeat the process until C1 is 0.

(ii) Finding smallest number:

e. Load the array count in a register C1.f. Get the first two numbers.g. Compare the numbers and exchange if the number is large.h. Get the third number from the array and repeat the process until C1 is 0.




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FLOWCHART

LARGEST NUMBER IN AN ARRAY SMALLEST NUMBER IN AN ARRAY

MAX = POINTER

IS MAX

INITIALIZECOUNT

PONITER =

POINTER + 1

COUNT = COUNT-1

STORE MAXIMUM

IS COUNT = 0?

YES

NO

YES

NO

STOP

START START

INITIALIZECOUNT

PONITER =

POINTER + 1

IS MIN

MIN = POINTER

COUNT = COUNT-1

IS COUNT = 0?

STORE MINIIMUM

STOP

YES

NO

NO

YES




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LARGEST


MOV SI,1200H Initialize array size

MOV CL,[SI] Initialize the count

INC SI Go to next memory location

MOV AL,[SI] Move the first data in AL

DEC CL Reduce the count

L2 : INC SI Move the SI pointer to next data

CMP AL,[SI] Compare two datas

JNB L1 If AL > [SI] then go to L1 ( no swap)

MOV AL,[SI] Else move the large number to AL

L1 : DEC CL Decrement the count

JNZ L2 If count is not zero go to L2

MOV DI,1300H Initialize DI with 1300H

MOV [DI],AL Else store the biggest number in 1300location

HLT Stop

SMALLEST


MOV SI,1200H Initialize array size

MOV CL,[SI] Initialize the count

INC SI Go to next memory location

MOV AL,[SI] Move the first data in AL

DEC CL Reduce the count

L2 : INC SI Move the SI pointer to next data

CMP AL,[SI] Compare two datas

JB L1 If AL < [SI] then go to L1 ( no swap)

MOV AL,[SI] Else move the large number to AL

L1 : DEC CL Decrement the count

JNZ L2 If count is not zero go to L2

MOV DI,1300H Initialize DI with 1300H

MOV [DI],AL Else store the biggest number in 1300location

HLT Stop




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RESULT:.

LARGEST

MEMORY

DATA

SMALLEST

MEMORY

DATA

Thus largest and smallest number is found in a given array.




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EXPT NO:05 DATE:

AIM:To write an Assembly Language Program (ALP) for performing the Arithmetic

operation of two byte numbers

APPARATUS REQUIRED:

SL.NO


1. Microprocessor kit 8086 kit 1


PROGRAM:;PASSWORD IS MASM1234DATA SEGMENTPASSWORD DB 'MASM1234'LEN EQU ($-PASSWORD)MSG1 DB 10,13,'ENTER YOUR PASSWORD: $'MSG2 DB 10,13,'WELCOME TO ELECTRONICS WORLD!!$'MSG3 DB 10,13,'INCORRECT PASSWORD!$'

NEW DB 10,13,'$'INST DB 10 DUP(0)DATA ENDSCODE SEGMENTASSUME CS:CODE,DS:DATASTART:MOV AX,DATAMOV DS,AXLEA DX,MSG1MOV AH,09HINT 21H

MOV SI,00UP1:MOV AH,08HINT 21HCMP AL,0DHJE DOWNMOV [INST+SI],ALMOV DL,'*'MOV AH,02HINT 21HINC SIJMP UP1DOWN:

PASSWORD CHECKING, PRINT RAM SIZE AND SYSTEM DATE




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MOV BX,00MOV CX,LENCHECK:MOV AL,[INST+BX]MOV DL,[PASSWORD+BX]

CMP AL,DLJNE FAILINC BXLOOP CHECKLEA DX,MSG2MOV AH,09HINT 21HJMP FINISHFAIL:LEA DX,MSG3MOV AH,009H

INT 21HFINISH:INT 3CODE ENDSEND STARTEND

;Today.asm Display month/day/year.; Feb 1st, 2012;CIS 206 Ken Howard

.MODEL small

.STACK 100h

.DATAmess1 DB 10, 13, 'Today is $' ; 10=LF, 13=CR.CODE

Today PROCMOV AX, @data

MOV DS, AXMOV DX, OFFSET mess1 ; Move string to DX

MOV AH, 09h ; 09h call to display string (DX > AH > DOS)

INT 21H ; Send to DOS; CX year, DH month, DL dayMOV AH, 2AH ; Get the date (appendix D)INT 21H ; Send to DOSPUSH CX ; Move year to the stack

MOV CX, 0 ; Clear CX

MOV CL, DLPUSH CX ; Move day to stackMOV CL, DH ; Move month > CL

PUSH CX ; Move month to stack

MOV DH, 0 ; Clear DH; ************************** DISPLAY MONTH ************************




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; Set up for division; Dividend will be in DX/AX pair (4 bytes)

; Quotient will be in AX; Remainder will be in DX

MOV DX, 0 ; Clear DX

POP AX ; Remove month from stack into AXMOV CX, 0 ; Initialize the counter

MOV BX, 10 ; Set up the divisordividem:

DIV BX ; Divide (will be word sized)PUSH DX ; Save remainder to stackADD CX, 1 ; Add one to counter

MOV DX, 0 ; Clear the remainderCMP AX, 0 ; Compare quotient to zeroJNE dividem ; If quoient is not zero, go to "dividem:"

divdispm:

POP DX ; Remove top of stack into DXADD DL, 30h ; ADD 30h (2) to DL

MOV AH, 02h ; 02h to display AH (DL)INT 21H ; Send to DOSLOOP divdispm ; If more to do, divdispm again

; LOOP subtracts 1 from CX. If non-zero, loop.MOV DL, '/' ; Character to display goes in DL

MOV AH, 02h ; 02h to display AH (DL)INT 21H ; Send to DOS

; ************************** DISPLAY DAY ************************; Set up for division

; Dividend will be in DX/AX pair (4 bytes); Quotient will be in AX; Remainder will be in DX

MOV DX, 0 ; Clear DXPOP AX ; Remove day from stack into AXMOV CX, 0 ; Initialize the counterMOV BX, 10 ; Set up the divisor

divided:DIV BX ; Divide (will be word sized)

PUSH DX ; Save remainder to stack

ADD CX, 1 ; Add one to counterMOV DX, 0 ; Clear the remainderCMP AX, 0 ; Compare quotient to zeroJNE divided ; If quoient is not zero, go to "divided:"

divdispd:POP DX ; Remove top of stack

ADD DL, 30h ; ADD 30h (2) to DLMOV AH, 02h ; 02h to display AH (DL)INT 21H ; Send to DOS

LOOP divdispd ; If more to do, divdispd again; LOOP subtracts 1 from CX. If non-zero, loop.




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MOV DL, '/' ; Character to display goes in DLMOV AH, 02h ; 02h to display AH (DL)INT 21H ; Send to DOS

; ************************** DISPLAY YEAR ************************; Set up for division

; Dividend will be in DX/AX pair (4 bytes); Quotient will be in AX; Remainder will be in DX

MOV DX, 0 ; Clear DXPOP AX ; Remove month from stack into AXMOV CX, 0 ; Initialize the counter

MOV BX, 10 ; Set up the divisordividey:

DIV BX ; Divide (will be word sized)PUSH DX ; Save remainder to stackADD CX, 1 ; Add one to counter

MOV DX, 0 ; Clear the remainderCMP AX, 0 ; Compare quotient to zeroJNE dividey ; If quoient is not zero, go to "dividey:"

divdispy:POP DX ; Remove top of stack into DX

ADD DL, 30h ; ADD 30h (2) to DLMOV AH, 02h ; 02h to display AH (DL)INT 21H ; Send to DOS

LOOP divdispy ; If more to do, divdisp again; LOOP subtracts 1 from CX. If non-zero, loop.

MOV al, 0 ; Use 0 as return codeMOV AH, 4ch ; Send return code to AHINT 21H ; Send return code to DOS to exit.

Today ENDP ; End procedureEND Today ; End code. Start using "Today" procedure.MVI A, 80H: Initialize 8255, port A and port BOUT 83H (CR): in output modeSTART: MVI A, 09HOUT 80H (PA): Send data on PA to glow R1 and R2MVI A, 24H

OUT 81H (PB): Send data on PB to glow G3 and G4MVI C, 28H: Load multiplier count (40) for delayCALL DELAY: Call delay subroutineMVI A, 12HOUT (81H) PA: Send data on Port A to glow Y1 and Y2OUT (81H) PB: Send data on port B to glow Y3 and Y4MVI C, 0AH: Load multiplier count (10) for delayCALL: DELAY: Call delay subroutineMVI A, 24HOUT (80H) PA: Send data on port A to glow G1 and G2MVI A, 09H

OUT (81H) PB: Send data on port B to glow R3 and R4MVI C, 28H: Load multiplier count (40) for delay




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CALL DELAY: Call delay subroutineMVI A, 12HOUT PA: Send data on port A to glow Y1 and Y2OUT PB: Send data on port B to glow Y3 and Y4MVI C, 0AH: Load multiplier count (10) for delay

CALL DELAY: Call delay subroutineJMP STARTDelay Subroutine:DELAY: LXI D, Count: Load count to give 0.5 sec delayBACK: DCX D: Decrement counterMOV A, DORA E: Check whether count is 0JNZ BACK: If not zero, repeatDCR C: Check if multiplier zero, otherwise repeatJNZ DELAYRET: Return to main program

Ram size:

ORG 0000HCLR PSW3CLR PSW4

CPL AADD A, #01HMOV A,R3

AGAIN: SJMP AGAIN

RESULT:Thus the output for the Password checking, Print RAM size and system date was

executed successfully




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EXP.NO: 06 DATE:

AIM:To write an assembly language program in 8086 to Counters and Time Delay

APPARATUS REQUIRED:SL.NO ITEM SPECIFICATION QUANTITY


2. Power Supply +5 V, dc,+12 V dc 1

3. Stepper Motor Interface board - 1

4. Stepper Motor - 1

PROGRAM:.MODEL SMALL.DATA

MSGIN DB 'Enter delay duration (0-50): $'MSG1 DB 'This is Microprocessor!$'DELAYTIME DW 0000H

.CODE

MOV DX,@DATAMOV DS,DXLEA DX,MSGINMOV AH,09HINT 21H

IN1:MOV AH,01HINT 21HCMP AL,0DH ;

JE NXTSUB AL,30HMOV DL,ALMOV AX,BXMOV CL,0AHMUL CLMOV BX,AXAND DX,00FFHADD BX,DXMOV DELAYTIME,BXLOOP IN1

Counters and Time Delay




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NXT: MOV CX,DELAYTIMEMOV DL,10MOV AH,02HINT 21H

LEA SI,MSG1

LP: PUSH DXMOV DL,[SI]CMP DL,'$'JE NXT2

MOV AH,02HINT 21HADD SI,1POP DXMOV DI,DELAYTIME

MOV AH, 0INT 1AhMOV BX, DX

Delay:MOV AH, 0INT 1AhSUB DX, BXCMP DI, DXJA Delay

LOOP LP

NXT2: MOV AH,4CHINT 21H

END

RESULT:Thus the output for the Counters and Time Delay was executed successfully




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EXP.NO: 07 DATE:

AIM:To write an assembly language program in 8086 to Traffic light control

APPARATUS REQUIRED:

SL.NO ITEM SPECIFICATION QUANTITY



3. Traffic light control Interfaceboard

- 1

PROGRAM:

TRAFFIC LIGHT CONTROL




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EXP.NO: 08 DATE:

AIM:To write an assembly language program in 8086 to rotate the motor at different speeds.

APPARATUS REQUIRED:

SL.NO ITEM SPECIFICATION QUANTITY



3. Stepper Motor Interface board - 1

4. Stepper Motor - 1

PROBLEM STATEMENT:

Write a code for achieving a specific angle of rotation in a given time and particularnumber of rotations in a specific time.

THEORY:

A motor in which the rotor is able to assume only discrete stationary angularposition is a stepper motor. The rotary motion occurs in a stepwise manner from one

equilibrium position to the next.Two-phase scheme: Any two adjacent stator windings areenergized. There are two magnetic fields active in quadrature and none of the rotor pole facescan be in direct alignment with the stator poles. A partial but symmetric alignment of therotor poles is of course possible.

ALGORITHM:

For running stepper motor clockwise and anticlockwise directions(i)

Get the first data from the lookup table.(ii) Initialize the counter and move data into accumulator.(iii)

Drive the stepper motor circuitry and introduce delay

(iv)

Decrement the counter is not zero repeat from step(iii)(v)

Repeat the above procedure both for backward and forward directions.

SWITCHING SEQUENCE OF STEPPER MOTOR:

MEMORY

LOCATION

A1 A2 B1 B2 HEX

CODE

4500 1 0 0 0 09 H

4501 0 1 0 1 05 H

4502 0 1 1 0 06 H

4503 1 0 1 0 0A H

STEPPER MOTOR INTERFACING




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FLOWCHART:

PROGRAM TABLE

ADDRESS OPCODE PROGRAM COMMENTS

START : MOV DI, 1200H Initialize memory location to store thearray of number

MOV CX, 0004H Initialize array size

LOOP 1 : MOV AL,[DI] Copy the first data in AL

OUT 0C0,AL Send it through port address

MOV DX, 1010H Introduce delayL1 : DEC DX

JNZ L1

INC DI Go to next memory location

LOOP LOOP1 Loop until all the datas have been sent

JMP START Go to start location for continuousrotation

1200 : 09,05,06,0A Array of datas

RESULT: Thus the assembly language program for rotating stepper motor in both clockwiseand anticlockwise directions is written and verified.

START

INTIALIZE COUNTER FOR LOOK UP TABLE

GET THE FIRST DATA FROM THE ACCUMULATOR

MOVE DATA INTO THE ACCUMULATOR

DRIVE THE MOTOR

DECREMENT COUNTER

GET THE DATA FROM LOOK UP

IS B = 0 ?

DELAY




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EXP.NO: 09 DATE:

AimTo display the digital clock specifically by displaying the hours, minutes and seconds

using 8086 kits

Apparatus required

S.No Item Specification

1 Microprocessor kit 8086

2 Power Supply 5V

Preliminary SettingsOrg 1000h

Store time value in memory location 1500- Seconds1501- Minutes1502- Hours

Digital clock program

DIGITAL CLOCK




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ResultThus the digital clock program has been written and executed using 8086

microprocessor kit and the output of digital clock was displayed as [hours: minutes: seconds]successfully.




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EXP.NO:10 DATE:

AIM :To display the rolling message HELP US in the display.

APPARATUS REQUIRED:8086 Microprocessor kit, Power supply, Interfacing board.

ALGORITHM :Display of rolling message HELP US

1.

Initialize the counter2. Set 8279 for 8 digit character display, right entry3.

Set 8279 for clearing the display4. Write the command to display5.

Load the character into accumulator and display it6. Introduce the delay7.

Repeat from step 1.

1. Display Mode Setup: Control word-10 H

0 0 0 1 0 0 0 0

0 0 0 D D K K K

DD00- 8Bit character display left entry01- 16Bit character display left entry10- 8Bit character display right entry11- 16Bit character display right entry

KKK- Key Board Mode000-2Key lockout.

2.Clear Display:Control word-DC H

1 1 0 1 1 1 0 0

1 1 0 CD CD CD CF CA

11 A0-3; B0-3 =FF

INTERFACING PRGRAMMABLE KEYBOARD AND

DISPLAY CONTROLLER- 8279

1-Enables Clear display

0-Contents of RAM will be displayed

1-FIFO Status is cleared

1-Clear all bits

(Combined effect of CD)




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3. Write Display: Control word-90H

1 0 0 1 0 0 0 0

1 0 0AI A A A A

FLOWCHART

SEGMENT DEFINITION

DATA BUS D7 D6 D5 D4 D3 D2 D1 D0

SEGMENTS d c b a dp g f e

Selects one of the 16 rows of display.

Auto increment = 1, the row address selected will be incremented after each of read andwrite operation of the display RAM.




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PROGRAM TABLE

PROGRAM COMMENTS

START : MOV SI,1200H Initialize array

MOV CX,000FH Initialize array size

MOV AL,10 Store the control word for display mode

OUT C2,AL Send through output port

MOV AL,CC Store the control word to clear display


MOV AL,90 Store the control word to write display


L1 : MOV AL,[SI] Get the first data


CALL DELAY Give delay

INC SI Go & get next data

LOOP L1 Loop until all the datas have been taken

JMP START Go to starting location

DELAY : MOV DX,0A0FFH Store 16bit count value

LOOP1 : DEC DX Decrement count value

JNZ LOOP1 Loop until count values becomes zero

RET Return to main program

LOOK-UP TABLE:

1200 98 68 7C C8

1204 FF 1C 29 FF

RESULT:

MEMORY

LOCATION

7-SEGMENT LED FORMAT HEX DATA

d c b a dp e g f

1200H 1 0 0 1 1 0 0 0 98

1201H 0 1 1 0 1 0 0 0 68

1202H 0 1 1 1 1 1 0 0 7C

1203H 1 1 0 0 1 0 0 0 C8

1204H 1 1 1 1 1 1 1 1 FF

1205H 0 0 0 0 1 1 0 0 1C

1206H 0 0 1 0 1 0 0 1 29

1207H 1 1 1 1 1 1 1 1 FF

Thus the rolling message HELP US is displayed using 8279 interface kit.




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EXP.NO:11 DATE:AIM:To display the Printer Status in the display

APPARATUS REQUIRED:8086 Microprocessor kit, Power supply, interfacing board.

PROGRAM:

XOR AX, AXXOR BX, BX

;this divides my 3digit number by 100 giving me my, hundredth digitMOV AX, RESMOV BX, 100DIV BX

;prints the hundredth digitADD AL, '0'MOV DL, ALPUSH AX ; save AX on the stackMOV AH, 02hINT 21hPOP AX ; restore ax

;divides the remainder by 10 giving me my tens digitMOV BX, 10DIV BX

;prints my tens digitADD AL, '0'MOV DL, AL

PUSH AX ; save AX on the stackMOV AH, 02hINT 21hPOP AX ; restore ax

;print my last remainder which is my onesADD AH, '0'MOV DL, AHMOV AH, 02hINT 21h

RESULT:Thus the output for the Move a data block without overlap was executed successfully

PRINTER STATUS




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EXP.NO: 12 DATE:

AimTo connect two 8086 microprocessor kits and to serially communicate with

each other by considering transmitter and receiver kits.

Apparatus required

Procedure

1. Take two no of 8086 microprocessor kits.

2. Enter the transmitter program in transmitter kit.

3. Enter the receiver program in receiver kit.

4. Interface the two kits with 9-9 serial cable in the serial port of the microprocessor kits.(LCD kit means PC-PC cable. LED kit means kit-kit cable)

5. Enter the data in transmitter kit use the memory location 1500.

6. Execute the receiver kit.

7. Execute the transmitter kit.

8. Result will be available in receiver kit memory location 1500.

Transmitter Program

SERIAL INTERFACE AND PARALLEL INTERFACE




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Receiver Program

Result

Thus the serial communication between two 8086 microprocessor kits has beenestablished and the data is transmitted in one kit and received in the other kitsuccessfully




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PARALLEL COMMUNICATION BETWEEN TWO 8086 MICROPROCESSORS KITS

AimTo connect two 8086 microprocessor kits and to establish parallel communication with eachother by considering transmitter and receiver kits.

Apparatus required

Procedure

1. Take two 8086 microprocessor kits.

2. Enter the transmitter program in transmitter kit.

3. Enter the receiver program in receiver kit.

4. Interface the two kits with 26-core cable on PPI-1.

5. Execute the receiver kit.

6. Execute the transmitter kit.

7. Go and see the memory location 1200 in receiver to get same eight data.

8. Data is available in transmitter kit in the memory location.

9. Change the data & execute the following procedure & get the result in receiver kit.

Transmitter program




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Receiver Program

ResultThus the serial communication between two 8086 microprocessor kits has been establishedand the data is transmitted in one kit and received in the other kit successfully.




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EXPT NO:13 DATE:AIM:

To write an assembly language program to convert an analog signal into a digital signalusing an ADC interfacing.

APPARATUS REQUIRED:

SL.NO ITEM SPECIFICATION QUANTITY1. Microprocessor kit 8086 1

2. Power Supply +5 V dc,+12 V dc 1

3. ADC Interface board - 1

THEORY:An ADC usually has two additional control lines: the SOC input to tell the ADC when to

start the conversion and the EOC output to announce when the conversion is complete. Thefollowing program initiates the conversion process, checks the EOC pin of ADC 0809 as towhether the conversion is over and then inputs the data to the processor. It also instructs the

processor to store the converted digital data at RAM location.

ALGORITHM:(i) Select the channel and latch the address.(ii) Send the start conversion pulse.(iii) Read EOC signal.(iv) If EOC = 1 continue else go to step (iii)(v) Read the digital output.

(vi)

Store it in a memory location.FLOW CHART:

START

SELECT THE CHANNEL AND LATCH

SEND THE START CONVERSION PULSE

READ THE DIGITALOUTPUT

STORE THE DIGITAL VALUE IN THE

MEMORY LOCATION SPECIFIED

IS EOC = 1?

STOP

NO

YES

A/D AND D/A INTERFACE AND WAVEFORM GENERATION




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PROGRAM TABLE

PROGRAM COMMENTS

MOV AL,00 Load accumulator with value for ALE high

OUT 0C8H,AL Send through output port

MOV AL,08 Load accumulator with value for ALE low

OUT 0C8H,AL Send through output port

MOV AL,01 Store the value to make SOC high in the accumulator

OUT 0D0H,AL Send through output port

MOV AL,00

Introduce delayMOV AL,00

MOV AL,00MOV AL,00 Store the value to make SOC low the accumulator

OUT 0D0H,AL Send through output port

L1 : IN AL, 0D8H

Read the EOC signal from port & check for end of

conversionAND AL,01

CMP AL,01

JNZ L1 If the conversion is not yet completed, read EOC signal

from port again

IN AL,0C0H Read data from port

MOV BX,1100 Initialize the memory location to store data

MOV [BX],AL Store the data

HLT Stop

RESULT:

ANALOG

VOLTAGE

DIGITAL DATA ON LED

DISPLAY

HEX CODE IN MEMORY

LOCATION

Thus the ADC was interfaced with 8086 and the given analog inputs were converted into itsdigital equivalent.




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INTERFACING DIGITALTOANALOG CONVERTER

AIM :1. To write an assembly language program for digital to analog conversion2. To convert digital inputs into analog outputs & To generate different waveforms

APPARATUS REQUIRED:

SL.NO ITEM SPECIFICATION QUANTITY1. Microprocessor kit 8086 Vi Microsystems 1


3. DAC Interface board - 1

PROBLEM STATEMENT:The program is executed for various digital values and equivalent analog voltages are

measured and also the waveforms are measured at the output ports using CRO.

THEORY:Since DAC 0800 is an 8 bit DAC and the output voltage variation is between 5v

and +5v. The output voltage varies in steps of 10/256 = 0.04 (approximately). The digital datainput and the corresponding output voltages are presented in the table. The basic idea behindthe generation of waveforms is the continuous generation of analog output of DAC. With 00(Hex) as input to DAC2 the analog output is 5v. Similarly with FF H as input, the output is+5v. Outputting digital data 00 and FF at regular intervals, to DAC2, results in a square waveof amplitude 5v.Output digital data from 00 to FF in constant steps of 01 to DAC2. Repeatthis sequence again and again. As a result a saw-tooth wave will be generated at DAC2output. Output digital data from 00 to FF in constant steps of 01 to DAC2. Output digital datafrom FF to 00 in constant steps of 01 to DAC2. Repeat this sequence again and again. As aresult a triangular wave will be generated at DAC2 output.

ALGORITHM:

Measurement of analog voltage:

(i)

Send the digital value of DAC.(ii)

Read the corresponding analog value of its output.

Waveform generation:Square Waveform:(i) Send low value (00) to the DAC.(ii)

Introduce suitable delay.(iii) Send high value to DAC.(iv) Introduce delay.(v) Repeat the above procedure.Saw-tooth waveform:(i)

Load low value (00) to accumulator.

(ii)

Send this value to DAC.(iii)

Increment the accumulator.(iv) Repeat step (ii) and (iii) until accumulator value reaches FF.(v)

Repeat the above procedure from step 1.Triangular waveform:(i)

Load the low value (00) in accumulator.(ii) Send this accumulator content to DAC.(iii) Increment the accumulator.(iv) Repeat step 2 and 3 until the accumulator reaches FF, decrement the

accumulator and send the accumulator contents to DAC.(v) Decrementing and sending the accumulator contents to DAC.

(vi)

The above procedure is repeated from step (i)




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FLOWCHART:MEASUREMENT OF ANALOG VOLTAGE SQUARE WAVE FORM

TRIANGULAR WAVEFORM

SAWTOOTH WAVEFORM

START

SEND THEDIGITALVALUE TO

ACCUMULATOR

TRANSFER THEACCUMULATOR

READ THE CORRESPONDINGANALOG VALUE

STOP

INTIALISE THEACCUMULATOR SEND ACC

LOAD THE ACC WITH MAXVALUE SEND ACC CONTENT

START

INITIALIZE

ACCUMULATOR

SEND ACCUMULATOR

INCREMENTACCUMULATOR

IS ACC

FF

YESNO

START

INITIALIZEACCUMULATOR

SEND ACCUMULATOR

CONTENT TO DAC

INCREMENTACCUMULATOR

DECREMENT

ACCUMULATOR CONTENT

SENDACCUMULATOR

START

DELAY

DELAY

IS ACCFF

NO

YES

IS ACC00YES NO




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MEASUREMENT OF ANALOG VOLTAGE:

PROGRAM COMMENTS

MOV AL,7FH Load digital value 00 in accumulator


HLT Stop

DIGITAL DATA ANALOG VOLTAGE

PROGRAM TABLE: Square Wave

PROGRAM COMMENTS

L2 : MOV AL,00H Load 00 in accumulator


CALL L1 Give a delay

MOV AL,FFH Load FF in accumulator


CALL L1 Give a delay

JMP L2 Go to starting location

L1 : MOV CX,05FFH Load count value in CX register

L3 : LOOP L3 Decrement until it reaches zero

RET Return to main program

PROGRAM TABLE: Saw tooth Wave

PROGRAM COMMENTS

L2 : MOV AL,00H Load 00 in accumulator

L1 : OUT C0,AL Send through output port

INC AL Increment contents of accumulator

JNZ L1 Send through output port until it reaches FF





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PROGRAM TABLE: Tri angular Wave

PROGRAM COMMENTS

L3 : MOV AL,00H Load 00 in accumulatorL1 : OUT C0,AL Send through output port

INC AL Increment contents of accumulator

JNZ L1 Send through output port until it reaches FF

MOV AL,0FFH Load FF in accumulator

L2 : OUT C0,AL Send through output port

DEC AL Decrement contents of accumulator

JNZ L2 Send through output port until it reaches 00


RESULT:WAVEFORM GENERATION:

WAVEFORMS AMPLITUDE TIMEPERIOD

Square Waveform

Saw-tooth waveform

Triangular waveform

MODEL GRAPH:

Square Waveform Saw-tooth waveform

Triangular waveform

Thus the DAC was interfaced with 8085 and different waveforms have been generated.




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EXPT NO:14 DATE:

8 BIT ADDITION

AIM:To write a program to add two 8-bit numbers using 8051 microcontroller.

ALGORITHM:

1. Clear Program Status Word.2. Select Register bank by giving proper values to RS1 & RS0 of PSW.3. Load accumulator A with any desired 8-bit data.4. Load the register R0with the second 8- bit data.5. Add these two 8-bit numbers.6. Store the result.7. Stop the program.

FLOW CHART:

START

Clear PSW

Select Register

Load A and R0with 8- bit datas

Add A & R0

Store the sum

STOP

BASIC ARITHMETIC AND LOGICAL OPERATIONS




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8 Bit Addition (Immediate Addressing)

ADDRESS LABEL MNEMONIC OPERAND HEX

CODE

COMMENTS

4100 CLR C C3 Clear CY Flag

4101 MOV A, data1 74,data1 Get the data1 inAccumulator

4103 ADDC A, # data 2 24,data2 Add the data1 with

data2

4105 MOV DPTR, #

4500H

90,45,00 Initialize the memory

location

4108 MOVX @ DPTR, A F0 Store the result in

memory location

4109 L1 SJMP L1 80,FE Stop the program

RESULT:

OUTPUT

MEMORY LOCATION DATA

4500

Thus the 8051 ALP for addition of two 8 bit numbers is executed.




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8 BIT SUBTRACTION

AIM:To perform subtraction of two 8 bit data and store the result in memory.

ALGORITHM:

a.

Clear the carry flag.b. Initialize the register for borrow.c.

Get the first operand into the accumulator.d. Subtract the second operand from the accumulator.e.

If a borrow results increment the carry register.f. Store the result in memory.

FLOWCHART:

START

CLEAR CARRY

FLAG

GET IST

OPERAND IN

ACCR

SUBTRACT THE

2ND OPERANDFROM ACCR

STORERESULT IN

MEMORY

STOP

IS CF=1

INCREMENTTHE BORROW

REGISTER

Y

N




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8 Bit Subtraction (Immediate Addressing)


CODE

COMMENTS

4100 CLR C C3 Clear CY flag

4101 MOV A, # data1 74, data1 Store data1 inaccumulator

4103 SUBB A, # data2 94,data2 Subtract data2 from

data1

4105 MOV DPTR, # 4500 90,45,00 Initialize memory

location

4108 MOVX @ DPTR, A F0 Store the difference

in memory location

4109 L1 SJMP L1 80,FE Stop

RESULT:

OUTPUTMEMORY LOCATION DATA

4500

Thus the 8051 ALP for subtraction of two 8 bit numbers is executed .




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8 Bit Multiplication


CODE

COMMENTS

4100 MOV A ,#data1 74, data1 Store data1 in

accumulator

4102 MOV B, #data2 75,data2 Store data2 in B reg

4104 MUL A,B F5,F0 Multiply both

4106 MOV DPTR, #

4500H

90,45,00 Initialize memory

location

4109 MOVX @ DPTR, A F0 Store lower order

result

401A INC DPTR A3 Go to next memory

location

410B MOV A,B E5,F0

Store higher order

result410D MOV @ DPTR, A F0

410E STOP SJMP STOP 80,FE Stop

RESULT:

INPUT OUTPUTMEMORY LOCATION DATA MEMORY LOCATION DATA

4500 4502

4501 4503

Thus the 8051 ALP for multiplication of two 8 bit numbers is executed .




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8 BIT DIVISION

AIM:To perform division of two 8 bit data and store the result in memory.

ALGORITHM:1. Get the Dividend in the accumulator.2.

Get the Divisor in the B register.3. Divide A by B.4.

Store the Quotient and Remainder in memory.

FLOWCHART:

START

GET DIVIDEND

IN ACCR

GET DIVISOR INB REG

DIVIDE A BY B

STOREQUOTIENT &REMAINDERIN MEMORY

STOP




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8 Bit Division


CODE

COMMENTS

4100 MOV A, # data1 74,data1 Store data1 in

accumulator

4102 MOV B, # data2 75,data2 Store data2 in B reg

4104 DIV A,B 84 Divide

4015 MOV DPTR, # 4500H 90,45,00 Initialize memory

location

4018 MOVX @ DPTR, A F0 Store remainder

4109 INC DPTR A3 Go to next memory

location

410A MOV A,B E5,F0

Store quotient

410C MOV @ DPTR, A F0

410D STOP SJMP STOP 80,FE Stop

RESULT:

INPUT OUTPUT

MEMORY LOCATION DATA MEMORY LOCATION DATA

4500 (dividend) 4502 (remainder)

4501 (divisor) 4503 (quotient)

Thus the 8051 ALP for division of two 8 bit numbers is executed .




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LOGICAL AND BIT MANIPULATION

AIM:

To write an ALP to perform logical and bit manipulation operations using 8051microcontroller.

APPARATUS REQUIRED:8051 microcontroller kit

ALGORITHM:

1. Initialize content of accumulator as FFH2. Set carry flag (cy = 1).3. AND bit 7 of accumulator with cy and store PSW format.4. OR bit 6 of PSW and store the PSW format.5. Set bit 5 of SCON.

6.

Clear bit 1 of SCON.7. Move SCON.1 to carry register.8.

Stop the execution of program.

FLOWCHART:

START

Set CY flag, AND CY with MSB of

Store the PSW format OR CY with bit 2 IE re

Clear bit 6 of PSW, Store PSW

Set bit 5 of SCON , clear bit 1 and storeSCON

Move bit 1 of SCON to CY and store PSW

STOP




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PROGRAM TABLE

ADDRESS HEX

CODE

LABEL MNEMONICS OPERAND COMMENT

4100 90,45,00 MOV DPTR,#4500 Initialize memorylocation

4103 74,FF MOV A,#FF Get the data in

accumulator

4105 D3 SETB C Set CY bit

4016 82,EF ANL C, ACC.7 Perform AND with 7t

bit of accumulator

4018 E5,D0 MOV A,DOH

Store the result410A F0 MOVX @DPTR,A

410B A3 INC DPTR Go to next location

410C 72,AA ORL C, IE.2 OR CY bit with 2nd

bit

if IE reg

410E C2,D6 CLR PSW.6 Clear 6t

bit of PSW

4110 E5,D0 MOV A,DOH

Store the result4112 F0 MOVX @DPTR,A

4113 A3 INC DPTR Go to next location

4114 D2,90 SETB SCON.5 Set 5th

of SCON reg

4116 C2,99 CLR SCON.1 Clear 1st bit of SCON

reg

4118 E5,98 MOV A,98H

Store the result411A F0 MOVX @DPTR,A

411B A3 INC DPTR Go to next location

411C A2,99 MOV C,SCON.1 Copy 1st bit of SCON

reg to CY flag

411E E5,D0 MOV A,DOHStore the result

4120 F0 MOVX @DPTR,A

4122 80,FE L2 SJMP L2 Stop




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RESULT:

MEMORY

LOCATION

SPECIAL FUNCTION REGISTER FORMAT BEFORE

EXECUTION

AFTER

EXECUTION

4500H (PSW) CY AC FO RS1 RS0 OV - P 00H 88H

4501H (PSW) CY AC FO RS1 RS0 OV - P 40H 88H

4502H (SCON) SM0 SM1 SM2 REN TB8 RB8 TI RI 0FH 20H

4503H (PSW) CY AC FO RS1 RS0 OV - P FFH 09H

Thus the bit manipulation operation is done in 8051 microcontroller.




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EXPT NO: 15 DATE:

AIM:To convert Square and Cube program, Find 2s complement of a numberusing 8051

micro controller

RESOURCES REQUIERED:

8051 microcontroller kit

Keyboard

Power supply

PROGRAM:

org 0000h; sets the program counter to 0000h

mov a,#n;assign value 'n' in decimal to A which is converted to it's

equivalent hexadecimal value

mov b,#n;assign value 'n' in decimal to B which is converted to it's


mov r0,#n;assign value 'n' in decimal to R0 which is converted to it's


mul ab; multiplying 'A' with 'B'

mov 40h,a; lower byte is stored in address 40h

mov 41h,b; higher byte is stored in address 41h

mov r1,a; move value of 'A' to R1

mov a,b; move value of 'B' to 'A'

mov b,r0; move value of R0 to b

mul ab; multiply 'A' and 'B'

mov b,a; lower byte obtained is moved from 'A' to 'B'

mov r2,b; move value of 'B' to R2

mov a,r1; move value of R1 to 'A'

mov b,r0; move value of R0 to 'B'

mul ab; multiplying 'A' and 'B'

mov 50h,a; Lower byte obtained is stored in address 50h

Square and Cube program, Find 2s complement of a number




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mov r3,b; higher byte obtained is stored in R3

mov a,r2; move value from R2 to 'A'

add a,r3; add value of 'A' with R3 and store the value in 'A'

mov b,a; move value from 'A' to 'B'

mov 51h,b; store value obtained in address 51h

end

SQUARE PGM USING 8051

ORG 00h

02 LJMP MAIN

03 DELAY:

04 ;MOV R0,#205 MOV TMOD, #01H

06 MOV TH0, #HIGH (-50000)

07 MOV TL0, #LOW (-50000)

08 SETB TR0

09 JNB TF0,

10 CLR TF0

12 ;DJNZ R0,DELAY

13 RET

14 MAIN:

15 MOV DPTR,#300H

16 MOV A,#0FFH

17 MOV P1,A

18 BACK:

19 LCALL DELAY

20 MOV A,P1

21 MOVC A,@A+DPTR

22 ;MOV P2,#00H23 ;LCALL DELAY

24 MOV P2,A

25 SJMP BACK

26 ORG 300H

27 XSQR_TABLE:

28 DB 0,1,4,9,16,25,36,49,64,81

29 END

Thus the Square and Cube program, Find 2s complement of a number is done in

8051 microcontroller




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EXPT NO:16 DATE:

AIM:To convert BCD number into ASCII by using 8051 micro controller

RESOURCES REQUIERED:

8051 microcontroller kit

Keyboard

Power supply

Unpacked BCD to ASCII




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FLOWCHART:


2.microprocessor microcontroller lab 1

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